Question

Evaluate \[ \lim_{n\to\infty}\frac{A+e^{nx}}{x+Ae^{nx}} \]

• A. \(\dfrac{A}{x}, \text{ when } x\lt 0\)
• B. \(1, \text{ when } x\gt 0\)
• C. \(0, \forall x\in\mathbb{R}\)
• D. \(A, \text{ when } x=0\)

Hint:

For limits involving \(e^{nx}\): \[ x\lt 0 \Rightarrow e^{nx}\to 0, \] and \[ x\gt 0 \Rightarrow e^{nx}\to \infty. \] Always analyse the limit separately for different signs of \(x\).

Answer 1

The Correct Option is A

Step 1: Identify the three cases for $x$.
The expression $\dfrac{A + e^{nx}}{x + Ae^{nx}}$ depends on whether $e^{nx}$ grows or shrinks as $n \to \infty$. This depends on the sign of $x$.

Step 2: Case $x > 0$.
If $x > 0$, then $nx \to +\infty$, so $e^{nx} \to \infty$. Divide numerator and denominator by $e^{nx}$: \[ \frac{Ae^{-nx} + 1}{xe^{-nx} + A} \to \frac{0 + 1}{0 + A} = \frac{1}{A}. \]

Step 3: Case $x < 0$.
If $x < 0$, then $nx \to -\infty$, so $e^{nx} \to 0$. The expression becomes: \[ \frac{A + e^{nx}}{x + Ae^{nx}} \to \frac{A + 0}{x + 0} = \frac{A}{x}. \]

Step 4: Case $x = 0$.
When $x = 0$: $e^{nx} = e^0 = 1$ for all $n$, so the expression $= \dfrac{A+1}{0+A} = \dfrac{A+1}{A}$, which is constant (not $A$ as stated in option 4).

Step 5: Match with the options.
Option 1 says the limit is $A/x$ when $x < 0$. This matches our Case 2 result.

Step 6: State the answer.
For $x < 0$, $\lim_{n\to\infty} \dfrac{A + e^{nx}}{x + Ae^{nx}} = \dfrac{A}{x}$.
\[ \boxed{\dfrac{A}{x} \text{ when } x < 0} \]